Paper Sh:566, version 1996-05-04 10. See https://shelah.logic.at/papers/566/ for possible updates.

A COMPLETE BOOLEAN ALGEBRA THAT HAS NO PROPER ATOMLESS COMPLETE SUBLAGEBRA

Thomas Jech and Saharon Shelah

The Pennsylvania State University, The Hebrew University and Rutgers University

Abstract. There exists a complete atomless Boolean algebra that has no proper atomless complete subalgebra.

An atomless complete Boolean algebra B is called simple [5] if it has no atomless complete subalgebra A such that A 6= B. We prove below that such an algebra exists.

The question whether a simple algebra exists was first raised in [8] where it was proved that B has no proper atomless complete subalgebra if and only if B is rigid

and minimal. For more on this problem, see [4], [5] and [1, p. 664].

Properties of complete Boolean algebras correspond to properties of generic mod- els obtained by forcing with these algebras. (See [6], pp. 266–270; we also follow [6] for notation and terminology of forcing and generic models.) When in [7] McAloon constructed a generic model with all sets ordinally definable he noted that the

corresponding complete Boolean algebra is rigid, i.e. admitting no nontrivial auto- morphisms. In [9] Sacks gave a forcing construction of a real number of minimal degree of constructibility. A complete Boolean algebra B that adjoins a minimal

The first author was supported in part by an NSF grant DMS-9401275. The second author was partially supported by the U.S.–Israel Binational Science Foundation. Publication No. 566

Typeset by AMS-TEX 1 Paper Sh:566, version 1996-05-04 10. See https://shelah.logic.at/papers/566/ for possible updates.

set (over the ground model) is minimal in the following sense:

(1) If A is a complete atomless subalgebra of B then there exists

a partition W of 1 such that for every w ∈ W , Aw = Bw,

where Aw = {a · w : a ∈ A}.

In [3], Jensen constructed, by forcing over L, a definable real number of minimal

degree. Jensen’s construction thus proves that in L there exists rigid minimal complete Boolean algebra. This has been noted in [8] and observed that B is rigid and minimal if and only if it has no proper atomless complete subalgebra. McAloon

then asked whether such an algebra can be constructed without the assumption that V = L. In [5] simple complete algebras are studied systematically, giving examples (in L) for all possible cardinalities.

In [10] Shelah introduced the (f, g)-bounding property of forcing and in [2] devel-

oped a method that modifies Sacks’ perfect tree forcing so that while one adjoins a minimal real, there remains enough freedom to control the (f, g)-bounding property. It is this method we use below to prove the following Theorem:

Theorem. There is a forcing notion P that adjoins a real number g minimal over

V and such that B(P) is rigid.

Corollary. There exists a countably generated simple complete Boolean algebra.

The forcing notion P consists of finitely branching perfect trees of height ω. In

order to control the growth of trees T ∈ P, we introduce a master tree T such that every T ∈ P will be a subtree of T . To define T , we use the following fast growing

∞ ∞ sequences of integers (Pk)k=0 and (Nk)k=0:

Pk (2) P0 = N0 = 1,Pk = N0 · ... · Nk−1,Nk = 2

211 (Hence Nk = 1, 2, 4, 256, 2 ,... ).

Definition. The master tree T and the index function ind:

(3)(i) T ⊂ [ω]<ω, 2 Paper Sh:566, version 1996-05-04 10. See https://shelah.logic.at/papers/566/ for possible updates.

(ii) ind is a one-to-one function of T onto ω, (iii) ind (< >) = 0, (iv) if s, t ∈ T and length(s) < length(t) then ind(s) < ind(t),

(v) if s, t ∈ T , length(s) = length(t) and s

(vi) if s ∈ T and ind(s) = k then s has exactly Nk successors in T , namely all

_ s i, i = 0,...,Nk − 1.

The forcing notion P is defined as follows:

Definition. P is the set of all subtrees T of T that satisfy the following:

(4) for every s ∈ T and every m there exists some t ∈ T , t ⊃ s,

m such that t has at least Pind(t) successors in T .

(We remark that T ∈ P because for every m there is a K such that for all k ≥ K,

m P Pk ≤ 2 k = Nk.) When we need to verify that some T is in P we find it convenient to replace (4) by an equivalent property:

Lemma. A tree T ⊆ T satisfies (4) if and only if

(5)(i) every s ∈ T has at least one successor in T , (ii) for every n, if ind(s) = n and s ∈ T then there exists a k such that if

n ind(t) = k then t ∈ T , t ⊃ s and t has at least Pk successors in T .

Proof. To see that (5) is sufficient, let s ∈ T and let m be arbitrary. Find some

s ∈ T such that s ⊃ s and ind(s) ≥ m, and apply (5ii). 

The forcing notion P is partially ordered by inclusion. A standard forcing ar- gument shows that if G is a generic subset of P then V [G] = V [g] where g is the

generic branch, i.e. the unique function g : ω → ω whose initial segments belong to all T ∈ G. We shall prove that the generic branch is minimal over V , and that the complete Boolean algebra B(P) admits no nontrivial automorphisms.

First we introduce some notation needed in the proof:

(6) For every k, sk is the unique s ∈ T such that ind(s) = k. 3 Paper Sh:566, version 1996-05-04 10. See https://shelah.logic.at/papers/566/ for possible updates.

(7) If T is a tree then s ∈ trunk(T ) if for all t ∈ T , either s ⊆ t or t ⊆ s.

(8) If T is a tree and a ∈ T then (T )a = {s ∈ T : s ⊆ a or a ⊆ s}.

Note that if T ∈ P and a ∈ T then (T )a ∈ P. We shall use repeatedly the following technique:

Lemma. Let T ∈ P and, let l be an integer and let U = T ∩ ωl (the lth level of

T ). Let x˙ be a name for some set in V . For each a ∈ U let Ta ⊆ (T )a and xa be

such that Ta ∈ P and Ta x˙ = xa. 0 S 0 0 l l Then T = {Ta : a ∈ U} is in P, T ⊆ T , T ∩ ω = T ∩ ω = U, and

0 T x˙ ∈ {xa : a ∈ U}. 

We shall combine this with fusion, in the form stated below:

∞ ∞ Lemma. Let (Tn)n=0 and (ln)n=0 be such that each Tn is in P, T0 ⊇ T1 ⊇ · · · ⊇

ln ln Tn ⊇ ... , l0 < l1 < ··· < ln < . . . , Tn+1 ∩ ω = Tn ∩ ω , and such that

(9) for every n, if sn ∈ Tn then there exists some t ∈ Tn+1, t ⊃ sn, with

n length(t) < ln+1, such that t has at least Pind(t) successors in Tn+1.

T∞ Then T = n=0 Tn ∈ P.

Proof. To see that T satisfies (5), note that if sn ∈ T then sn ∈ Tn, and the node

t found by (9) belongs to T . 

We shall now prove that the generic branch is minimal over V :

Lemma. If X ∈ V [G] is a set of ordinals, then either X ∈ V or g ∈ V [X].

Proof. The proof is very much like the proof for Sacks’ forcing. Let X˙ be a name for

X and let T0 ∈ P force that X˙ is not in the ground model. Hence for every T ≤ T0 0 00 0 ˙ 00 ˙ there exist T , T ≤ T and an ordinal α such that T α ∈ X and T α∈ / X. 0 0 Consequently, for any T1 ≤ T and T2 ≤ T there exist T1 ≤ T1 and T2 ≤ T2 and 0 0 ˙ 0 ˙ an α such that both T1 and T2 decide “α ∈ X” and T1 α ∈ X if and only if 0 ˙ T2 α∈ / X. 4 Paper Sh:566, version 1996-05-04 10. See https://shelah.logic.at/papers/566/ for possible updates.

∞ ∞ ln Inductively, we construct (Tn)n=0,(ln)n=0, Un = Tn ∩ ω , and ordinals α(a, b)

for all a, b ∈ Un, a 6= b, such that

(10)(i) Tn ∈ P and T0 ⊇ T1 ⊇ · · · ⊇ Tn ⊇ ... ,

(ii) l0 < l1 < ··· < ln < ··· ,

ln ln (iii) Tn+1 ∩ ω = Tn ∩ ω = Un,

(iv) for every n, if sn ∈ Tn then there exists some t ∈ Tn+1, t ⊃ sn, with

n length(t) < ln+1, such that t has at least Pind(t) successors in Tn+1,

(v) for every n, for all a, b ∈ Un, if a 6= b then both (Tn)a and (Tn)b decide ˙ ˙ ˙ “α(a, b) ∈ X” and (Tn)a α(a, b) ∈ X if and only if (Tn)b α(a, b) ∈ X. T∞ When such a sequence has been constructed, we let T = n=0 Tn. As (9) is

satisfied, we have T ∈ P and T ≤ T0. If G is a generic such that T ∈ G and if X is ˙ the G-interpretation of X then the generic branch g is in V [X]: for every n, g  ln

is the unique a ∈ Un with the property that for every b ∈ Un, b 6= a, α(a, b) ∈ X if ˙ and only if (T )a α(a, b) ∈ X. ∞ ∞ To construct (Tn)n=0,(ln)n=0 and α(a, b), we let l0 = 0 (hence U0 = {s0})

and proceed by induction. Having constructed Tn and ln, we first find ln+1 > ln

n as follows: If sn ∈ Tn, we find t ∈ Tn, t ⊃ sn, such that t has at least Pind(t)

successors in Tn. Let ln+1 = length (t) + 1. (If sn ∈/ Tn, let ln+1 = ln + 1.) Let

ln+1 Un+1 = Tn ∩ ω .

Next we consider, in succession, all pairs {a, b} of district elements of Un+1,

eventually constructing conditions Ta, a ∈ Un+1, and ordinals α(a, b), a, b ∈ Un+1, ˙ such that for all a, Ta ≤ (Tn)a and if a 6= b then either Ta α(a, b) ∈ X and ˙ ˙ ˙ Tb α(a, b) ∈/ X, or Ta α(a, b) ∈/ X and Tb α(a, b) ∈ X. Finally, we let S Tn+1 = {Ta : a ∈ Un+1}.

∞ ∞ It follows that (Tn)n=0,(ln)n=0 and α(a, b) satisfy (10). 

Let B be the complete Boolean algebra B(P). We shall prove that B is rigid. Toward a contradiction, assume that there exists an automorphism π of B that is

not the identity. First, there is some u ∈ B such that π(u) · u = 0. Let p ∈ P be such that p ≤ u and let q ∈ P be such that q ≤ π(p). Since q 6≤ p, there is some 5 Paper Sh:566, version 1996-05-04 10. See https://shelah.logic.at/papers/566/ for possible updates.

s ∈ q such that s∈ / p. Let T0 = (q)s.

Note that for all t ∈ T0, if t ⊇ s then t∈ / p. Let

A = {ind(t): t ∈ p},

and consider the following property ϕ(x) (with parameters in V): (11) ϕ(x) ↔ if x is a function from A into ω

such that x(k) < Nk for all k, then there exists

a function u on A in the ground model V such that the values of u are finite sets of

integers and for every k ∈ A, u(k) ⊆ {0, ..., Nk − 1} and |u(k)| ≤ Pk,

and x(k) ∈ u(k). We will show that

(12) p ∃x¬ϕ(x),

and

(13) there exists a T ≤ T0 such that T ∀xϕ(x).

This will yield a contradiction: the Boolean value of the sentence ∃x¬ϕ(x) is preserved by π, and so

T0 ≤ q ≤ π(p) ≤ π(k∃x¬ϕ(x)k) = k∃x¬ϕ(x)k,

contradicting (13). In order to prove (12), consider the following (name for a) functionx ˙ : A → ω. For every k ∈ A, let

x˙(k) =g ˙(length (sk) + 1) if sk ⊂ g,˙ andx ˙(k) = 0 otherwise.

Now if p1 < p and u ∈ V is a function on A such that u(k) ⊆ {0, ..., Nk − 1} and

|u(k)| ≤ Pk then there exist a p2 < p1 and some k ∈ A such that sk ∈ p2 has at

2 least Pk successors, and there exist in turn a p3 < p2 and some i∈ / u(k) such that

_ sk i ∈ trunk(p3). Clearly, p3 x˙(k) ∈/ u(k). Property (13) will follow from this lemma: 6 Paper Sh:566, version 1996-05-04 10. See https://shelah.logic.at/papers/566/ for possible updates.

Lemma. Let T1 ≤ T0 and x˙ be such that T1 forces that x˙ is function from A into ω

∞ ∞ ∞ such that x(k) < Nk for all k ∈ A. There exist sequences (Tn)n=1, (ln)n=1, (jn)n=1, ∞ (Un)n=1 and sets za, a ∈ Un, such that

(14)(i) Tn ∈ P and T1 ⊇ T2 ⊇ · · · ⊇ Tn ⊇ ... ,

(ii) l1 < l2 < ··· < ln < . . . ,

ln ln (iii) Tn+1 ∩ ω = Tn ∩ ω = Un,

(iv) for every n, if sn ∈ Tn then there exists some t ∈ Tn+1, t ⊃ sn, with

n length(t) < ln+1, such that t has at least Pind(t) successors in Tn+1,

(v) j1 < j2 < ··· < jn < . . . ,

(vi) for every a ∈ Un, (Tn)a hx˙(k): k ∈ A ∩ jni = za,

(vii) for every k ∈ A, if k ≥ jn then |Un| < Pk,

(viii) for every k ∈ A, if k < jn then |{za(k): a ∈ Un}| ≤ Pk.

T∞ Granted this lemma, (13) will follow: If we let T = n=1 Tn, then T ∈ P and

T ≤ T1 and for every k ∈ A, T x˙(k) ∈ u(k) where u(k) = {za(k): a ∈ Un} (for any and all n > k).

Proof of Lemma. We let l1 = j1 = length(s), U1 = {s} and strengthen T1 if

necessary so that T1 decides hx˙(k): k ∈ A ∩ j1i, and let zs be the decided value.

We also assume that length(s) ≥ 2 so that |U1| = 1 < Pk for every k ∈ A, k ≥ j1. Then we proceed by induction.

Having constructed Tn, ln, jn etc., we first find ln+1 > ln and jn+1 > jn as

follows: If sn ∈/ Tn (Case I), we let ln+1 = ln + 1 and jn+1 = jn + 1. Thus assume

that sn ∈ Tn (Case II).

Since length(sn) ≤ n ≤ ln, we choose some vn ∈ Un such that sn ⊆ vn. By (4)

n+1 there exists some t ∈ Tn, t ⊃ vn, so that if ind(t) = m then t has at least Pm

successors in Tn. Moreover we choose t so that m = ind(t) is big enough so that

there is at least one k ∈ A such that jn ≤ k < m. We let ln+1 = length(t) + 1 and

jn+1 = m = ind(t).

Next we construct Un+1, {za : a ∈ Un+1} and Tn+1. In Case I, we choose for

each u ∈ Un some successor a(u) of u and let Un+1 = {a(u): u ∈ Un}. For every 7 Paper Sh:566, version 1996-05-04 10. See https://shelah.logic.at/papers/566/ for possible updates.

a ∈ Un+1 we find some Ta ⊆ (Tn)a and za so that Ta hx˙(k): k ∈ A ∩ jn+1i = za, S and let Tn+1 = {Ta : a ∈ Un+1}. In this case |Un+1| = |Un| and so (vii) holds for n + 1 as well, while (viii) for n + 1 follow either from (viii) or from (vii) for n (the

latter if jn ∈ A).

Thus consider Case II. For each u ∈ Un other than vn we choose some a(u) ∈ Tn

of length ln+1 such that a(u) ⊃ u, and find some Ta(u) ⊆ (Tn)a(u) and za(u) so that

Ta(u) hx˙(k): k ∈ A ∩ mi = za(u). n+1 Let S be the set of all successors of t (which has been chosen so that |S| ≥ Pm

where m = ind(t)); every a ∈ S has length ln+1. For each a ∈ S we choose Ta ⊆

(Tn)a and za, so that Ta hx˙(k): k ∈ A ∩ mi = za. If we denote K = max(A ∩ m) then we have

K Y Y |{za : a ∈ S}| ≤ Ni ≤ Ni = PK+1 ≤ Pm, i∈A∩m i=0

n+1 n while |S| ≥ Pm . Therefore there exists a set U ⊂ S of size Pm such that for

every a ∈ U the set za is the same. Therefore if we let

Un+1 = U ∪ {a(u): u ∈ Un − {vn}},

S and Tn+1 = {Ta : a ∈ Un+1}, Tn+1 satisfies property (iv). It remains to verify that (vii) and (viii) hold.

To verify (vii), let k ∈ A be such that k ≥ jn+1 = m. Since m = ind(t), we have

m∈ / A and so k > m. Let K ∈ A be such that jn ≤ K < m. Since |Un| < PK , we have

|Un+1| < |Un| + |U| < PK + Nm < Pm · Nm = Pm+1 ≤ Pk.

To verify (viii), it suffices to consider only those k ∈ A such that jn ≤ k < m.

But then |Un| < Pk and we have

|{za(k): a ∈ Un+1}| ≤ |{za : a ∈ Un+1}| ≤ |Un| + 1 ≤ Pk.

 8 Paper Sh:566, version 1996-05-04 10. See https://shelah.logic.at/papers/566/ for possible updates.

References [1] M. Bekkali and R. Bonnet, Rigid Boolean algebras, in “Handbook of Boolean algebras” (J. D. Monk, ed.), vol. 2, pp. 637–678. [2] M. Goldstern and S. Shelah, Many simple cardinal invariants, Arch. Math. Logic 32 (1993), 203–221, (Publ. no. 448). [3] R. B. Jensen, Definable sets of minimal degree, in “Mathematical Logic and Foundations of Set Theory” (Y.Bar-Hillel, ed.), pp. 122–128, North-Holland Publ. Co., 1970. [4] T. Jech, A propos d’alg`ebres de Boole rigides et minimales, C. R. Acad. Sci. Paris S´er.A 274 (1972), 371–372. [5] T. Jech, Simple complete Boolean algebras, Israel J. Math. 18 (1974), 1–10. [6] T. Jech, Set Theory, Academic Press, New York 1978. [7] K. McAloon, Consistency results about ordinal definability, Ann. Math. Logic 2 (1971), 449– 467. [8] K. McAloon, Les alg`ebres de Boole rigides et minimales, C. R. Acad. Sci. Paris S´er.A 272 (1971), 89–91. [9] G. Sacks, Forcing with perfect closed sets, in “Axiomatic Set Theory” (D. Scott, ed.), Proc. Symp. Pure Math. 13, I, pp. 331–355, American Math. Society 1971. [10] S. Shelah, Vive la diff´erence I: nonisomorphism of ultrapowers of countable models, in “Set Theory of the Continuum” (H. Judah et al., eds.) MSRI vol. 26, pp.357–405, Springer-Verlag 1992.

Department of Mathematics, The Pennsylvania State University, University Park, PA 16802, USA School of Mathematics, The Hebrew University, Jerusalem, Israel, and Depart- ment of Mathematics, Rutgers University, New Brunswick, NJ 08903, USA E-mail address: [email protected], [email protected]

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